Integration: Antiderivatives
Rules of calculation for antiderivatives
As with differentiation, we can give a sum rule and constant rule for finding antiderivatives.
For functions #\blue f# and #\purple g#:
\[\begin{array}{c}
\displaystyle \int \blue{f(x)}+\purple{g(x)} \; {\dd}x = \displaystyle \int \blue{f(x)}\;{\dd}x + \int \purple{g(x)}\;{\dd}x
\end{array}\]
Examples
#\begin{array}{rcl}
\displaystyle \int \blue{2x} + \purple{4} \; {\dd}x &=& \displaystyle \int \blue{2x} \;{\dd}x + \int \purple{4} \;{\dd}x \\
&=& x^2 + 4x + \green {C}
\end{array}#
For a function #\blue f# and a constant #\orange c#:
\[\begin{array}{c}
\displaystyle \int \orange c \cdot \blue{f(x)} \; {\dd}x = \orange c\cdot \displaystyle \int \blue{f(x)}\;{\dd}x
\end{array}\]
Example
#\begin{array}{rcl}
\displaystyle \int \orange{9} \cdot \blue{x^2} \; \dd x &=& \orange{9} \cdot\displaystyle \int \blue{x^2} \; \dd x \\
&=& 9 \cdot \frac{1}{3}x^3 +\green C \\ &=& 3x^3+\green C
\end{array}#
#\begin{array}{rcl}
\displaystyle \int 2\cdot x^8+4\cdot x^6 \dd x &=&\displaystyle \int 2 \cdot x^{8}\, \dd x + \int 4 \cdot x^6\, \dd x\\
&&\phantom{xxx}\blue{\text{rule }\displaystyle \int f(x) + g(x) \, \dd = \int f(x) \, \dd x + \int g(x) \, \dd x}\\
&=&\displaystyle 2 \cdot \int x^{8}\, \dd x + 4 \cdot \int x^6 \, \dd x\\
&&\phantom{xxx}\displaystyle \blue{\text{twice applied the rule }\int c\cdot f(x)\,\dd x = c \cdot \int f(x) \, \dd x}\\
&=&\displaystyle \frac{2}{8+1}\cdot x^{8+1}+\frac{4}{6+1}\cdot x^{6+1}\\
&&\displaystyle \phantom{xxx}\blue{\text{twice applied the rule }\int x^{n} \; \dd x = \displaystyle\frac{1}{n+1} x^{n+1} + C}\\
&=&\displaystyle {{2\cdot x^9}\over{9}}+{{4\cdot x^7}\over{7}}+C\\
&&\displaystyle\phantom{xxx}\blue{\text{simplified}}
\end{array}#
Because we only asked for one antiderivative, we can now choose #C=0#. This gives:
\[F(x)={{2\cdot x^9}\over{9}}+{{4\cdot x^7}\over{7}}\]
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